# integral test to show that infinite product $\prod \limits_{n=1}^\infty\left(1+\frac{2}{n}\right)$ diverges

This is part of an assignment that I need to get a good mark for - I'd appreciate it if you guys could look over it and give some pointers where I've gone wrong.

(apologies for the italics)

$$\prod_{n=1}^\infty\left(1+\frac{2}{n}\right)\; \text{ converges when } \sum_1^\infty \ln\left(1+\frac{2}{n}\right)\; \text{ converges }.$$ $$\sum_1^\infty \ln\left(1+\frac{2}{n}\right)\;=\;\sum_1^\infty \ln(n+2)-\ln(n)$$ $$\text{let } f(x)=\ln(x+2)-\ln(x) \rightarrow f'(x)=\frac{1}{x+2} - \frac{1}{x}$$ $$= \frac{x-x-2}{x(x+2)} = \frac{-2}{x(x+2)}<0$$

$$f(x)\ \text{is a decreasing function}.$$ $$f(x) \; \text{is a positive function for} \;x\geq1$$ $$f(x)\;\text{is a continuous function for} \;x>=1$$

using integration test. $$\int_1^\infty \ln(x+2) - \ln(x) = \lim_{t \to \infty}\int_1^t \ln(x+2)dx - \lim_{t \to \infty}\int_1^t \ln x dx$$

$$\int \ln(x)dx = x \ln x - x + c \Rightarrow \int \ln(x+2) = (x+2)\ln(x+2) - (x+2) + c$$ Therefore $$\int \ln(x+2) - \ln(x)dx = (x+2)\ln(x+2)-x - 2 - x \ln(x) + x + c$$ $$= x \ln(\frac{x+2}{x})+ 2\ln(x+2)-2 + c$$ Therefore, $$\int_1^\infty \ln(x+2) - \ln(x)dx = \lim_{t \to \infty}\left[x \ln(\frac{x+2}{x}) + 2 \ln(x+2) - 2\right]_1^t$$

$$= \lim_{t \to \infty}\left[t \ln(\frac{t+2}{t}) + 2\ln(t+2) - 2\right] - \lim_{t \to \infty}\left[\ln(\frac{3}{1}) + 2\ln(3) - 2\right]$$

$$=\lim_{t \to \infty}\left[t \ln(\frac{t+2}{t}) + 2\ln(t+2) - 3\ln(3)\right]$$

$$As\; t\rightarrow\infty, \; \lim_{t \to \infty}t \ln\left(\frac{t+2}{t}\right) + 2\ln(t+2) = \infty.$$

Therefore the series $$\sum_1^\infty \ln\left(1+\frac{2}{n}\right)$$ is divergent.

Similarly the infinite product $$\prod_{n=1}^\infty\left(1+\frac{2}{n}\right)$$ is also divergent.

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You can put text that you don't want in math format in \mtext (or, even better, just take it out of math mode). –  Adrian Petrescu Dec 23 '10 at 5:11
Silly question: Is the answer wrong (that the product diverges), or is the proof wrong? Or neither? –  Jesse Madnick Dec 23 '10 at 5:15
Because that series really should diverge by telescoping... –  Jesse Madnick Dec 23 '10 at 5:16
I think he wants us to show the divergence by applying the integral test. –  gnicezw Dec 23 '10 at 5:25
The comment didn't offend, it is just irrelevant and obstructs the more relevant parts. E.g. the excerpt of the question on the main site reveals "Hi Guys, This is part of an assignment that I need to get a good mark for - I'd appreciate it if you guys could look over it and give some pointers where I've gone wrong. (apologies for the". Instead, it should be telling the users what the question is about (luckily, in this particular instance, your title already does a good job). –  Alex B. Dec 23 '10 at 8:52

For $a_n \ge 0$ the infinite product $\prod_{n=1}^\infty (1+ a_n)$ converges precisely when the infinite sum $\sum_{n=1}^\infty a_n$ converges, since

$$1+ \sum_{n=1}^N a_n \le \prod_{n=1}^N (1+ a_n) \le \exp \left( \sum_{n=1}^N a_n \right) .$$

So you only need consider $\sum_{n=1}^\infty 2/n$ and you can use your integral test for that, or just quote the standard result.

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If You dont have to use integration test then You can do something like this, it is much easier. $$\prod\limits_{n=1}^{k}(1+\frac{2}{n})=\frac{(k+1)(k+2)}{2}.$$
It is very easy to show since it is equal to: $\frac{1}{3}\cdot\frac{2}{4}\cdot\frac{3}{5}\cdot\frac{4}{6}\cdot\frac{5}{7}\cdot\cdot\cdot\cdot$
You can reduce almost all this fractions. Denominator of nth fraction can be reduced with numerator of n+2th fraction.

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